Evaluating the series $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}}$

Wolfram MathWorld states that $$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3) \, ,$$

where $\psi_{1}(x)$ is the trigamma function.

I can’t seem to get the answer in that form.

Using the Taylor expansion $\displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$, I get

$$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, dx.$$

Then I integrating by parts, I get

\begin{align} &4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \, dx \\ &= \frac{\pi^{2}}{9} \ln \left(\frac{1}{2} \right) – 8 \int_{0}^{\frac{1}{2}} \frac{\arcsin (x) \ln (x)}{\sqrt{1-x^{2}}} \, dx \\ &= – \frac{\pi^{2}}{9} \ln 2 – 8 \int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, du \\ &= – 8 \ln 2 \int_{0}^{\frac{\pi}{6}} u \, du – 8\int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \, du \\ &= – 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \, du \\ &= -8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \ln (1-e^{2iu}) \, du \\ &= 8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \sum_{n=1}^{\infty} \frac{e^{2in u}}{n} \, du \\ &= 8 \ \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\frac{\pi}{6}} u \cos (2nu) \, du \\ &= \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} – 2 \zeta(3) \\ &= \frac{2 \pi}{3} \Bigg( \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{2}} + \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{2}} – \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{2}} – \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{2}} \Bigg) \\ &+ \ 2 \Bigg( \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{3}} – \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{3}} – \sum_{n=0}^{\infty} \frac{1}{(6n+3)^{3}} – \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{3}} \\ &+ \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{3}} + \sum_{n=1}^{\infty} \frac{1}{(6n)^{3}} \Bigg) – 2 \zeta(3) \\ &= \frac{\pi \sqrt{3}}{108} \Bigg( \psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) – \psi_{1}\left(\frac{5}{6} \right) \Bigg) + \frac{1}{432} \Bigg( – \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &-28 \zeta(3) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) + 4 \zeta(3)\Bigg) – 2 \zeta (3) .\end{align}

From here I’ve been going in circles trying to get the answer in the form given above.

EDIT:

Using the duplication formula for the trigamma function (i.e., $\displaystyle 4 \psi_{1}(2x) = \psi_{1}(x) + \psi_{1} \left(x + \frac{1}{2} \right)$),

\begin{align} &\psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) – \psi_{1}\left(\frac{5}{6} \right) \\ &= 4 \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) – 4 \psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) \\ &= 6 \psi_{1} \left(\frac{1}{3} \right) – 6 \psi_{1} \left(\frac{2}{3} \right). \end{align}

Therefore,

\begin{align} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &= \frac{\sqrt{3} \pi}{18} \Bigg( \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Bigg) + \frac{1}{432} \Bigg( – \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &+ \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) – 2 \zeta(3)\Bigg) – 2 \zeta (3). \end{align}

So it comes down to somehow showing that $$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3) .$$

SECOND EDIT:

Using the duplication formula for $\psi_{2}(x)$, we get

\begin{align} &-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) – \psi_{2} \left(\frac{5}{6} \right) \\ &= -8 \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) – 8 \psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) \\ &= -6 \psi_{2} \left(\frac{1}{3} \right) – 6 \psi_{2} \left(\frac{2}{3} \right) . \end{align}

And using the more general multiplication formula, we get

$$\psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) + \psi_{2}(1) = 27 \psi_{2} (1) .$$

Therefore,

$$-6 \psi_{2} \left(\frac{1}{3} \right) – 6 \psi_{2} \left(\frac{2}{3} \right) = -156 \psi_{2} (1) = 312 \zeta(3) .$$

Solutions Collecting From Web of "Evaluating the series $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}}$"

Using the Taylor expansion for $\arcsin^2(x)$ and integrating, it’s easy to show (as you made) that
\begin{align} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &= 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \operatorname{d}x = – 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \operatorname{d}u \\ &= \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} – 2 \zeta(3)\\ &=\frac{2 \pi}{3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+2\operatorname{Cl}_3\left(\frac{\pi}{3}\right)-2 \zeta(3) \end{align}
recalling that the Clausen function are defined as
$$\operatorname{Cl}_{m}(\theta)= \begin{cases}\displaystyle \sum_{n=1}^{\infty} \frac{\sin (n\theta)}{n^{m}}& \text{for } m\text{ even}\\ \displaystyle\sum_{n=1}^{\infty} \frac{\cos(n\theta)}{n^{m}}& \text{for } m\text{ odd} \end{cases}$$
The value of the Clausen function $\operatorname{Cl}_3$ at $\frac{\pi}{3}$ is
$$\operatorname{Cl}_3\left(\frac{\pi}{3}\right)=\frac{1}{2}\left(1-2^{-2}\right)\left(1-3^{-2}\right)\zeta(3)=\frac{1}{3}\zeta(3)$$
From the duplication formula
$$\operatorname{Cl}_{2m}(2\theta)=2^{2m-1}\left[\operatorname{Cl}_{2m}(\theta)-\operatorname{Cl}_{2m}(\pi-\theta)\right]$$
we find
$$\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)=\frac{2}{3}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right).$$
From the identities for the trigamma function at $1/3$ and $2/3$
\begin{align} \psi_1\left(\frac{1}{3}\right) &=\frac{2\pi^2}{3}+3\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)\\ \psi_1\left(\frac{2}{3}\right) &=\frac{2\pi^2}{3}-3\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right) \end{align}
one has
$$\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)=6\sqrt 3\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)$$
and then
$$\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)=\frac{\sqrt 3}{12}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]$$

Finally, putting all together, we have
\begin{align} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &=\frac{2 \pi}{3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+2\operatorname{Cl}_3\left(\frac{\pi}{3}\right)-2 \zeta(3)\\ &=\frac{2 \pi}{3}\frac{\sqrt 3}{12}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]+2\frac{1}{3}\zeta(3)-2\zeta(3)\\ &=\frac{\pi\sqrt 3}{18}\left[\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)\right]-\frac{4}{3}\zeta(3). \end{align}

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) – \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3)}$

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n{2n \choose n}}
=\sum_{n = 1}^{\infty}x^{n}\,
{\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
=\sum_{n = 1}^{\infty}x^{n}\int_{0}^{1}t^{n – 1}\pars{1 – t}^{n}\,\dd t
\\[3mm]&=\int_{0}^{1}\sum_{n = 1}^{\infty}\bracks{tx\pars{1 – t}}^{n}
\,{\dd t \over t}
=\int_{0}^{1}\bracks{{1 \over 1 – tx\pars{1 – t}} – 1}\,{\dd t \over t}
=\int_{0}^{1}{x\pars{1 – t} \over 1 – tx\pars{1 – t}}\,\dd t
\end{align}

\begin{align}
&\sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}}
=\int_{0}^{x}\dd y\int_{0}^{1}{1 – t \over 1 – ty\pars{1 – t}}\,\dd t
=-\int_{0}^{1}{\ln\pars{1 – \bracks{1 – t}tx} \over t}\,\dd t
\end{align}

\begin{align}
&\color{#66f}{\large\sum_{n = 1}^{\infty}{1 \over n^{3}{2n \choose n}}}
=-\int_{0}^{1}{\dd x \over x}
\int_{0}^{1}{\ln\pars{1 – \bracks{1 – t}tx} \over t}\,\dd t
=\int_{0}^{1}{{\rm Li}_{2}\pars{t\bracks{1 – t}} \over t}\,\dd t
\\[3mm]&=\color{#66f}{{\pi \over 36\root{3}}\, \left[\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right]-\frac{4 \zeta (3)}{3}}
\approx {\tt 0.5229}
\end{align}